# Nash equilibrium two players in a joint effortful project

I have a problem where there are two agents in a joint project, Each agent $$i$$ puts in effort $$x_i$$ $$(0 \leq x_i \leq 1)$$ which cost each $$c(x_i)= x_i^2$$. The outcome of the project is $$f(x_1, x_2)= 3 x_1 x_2$$

which is split equally between both, regardless of their effort levels. I am asked to formulate de situation as a normal form game and find the NE.

What I have done is that I have assumed some values for the efforts to evaluate the game and I have also maximized each player's payoff individually to get best responses and make them intersect to get to the NE.

1. In this illustration, NE are $$(0,0)$$; $$(0.25, 0.25)$$ and $$(0.5, 0.5)$$

2. $$\max f(x_1, x_2)-c_1$$ with respect to $$x_1$$ ; yields : $$x_1 = (3/4)x_2$$ and $$\max f(x_1, x_2)-c_2$$ with respect to $$x_2$$ ; yields : $$x2 = (3/4)x_1$$ Both maximization problems intersect when $$x_1 = 0 = x_2$$

Anyhow, I don't think this is the right way to go...

You are actually almost correct. Your first approach shows the normal form of the game, but in an incomplete way, which is why you find "extra" Nash equilibria. The possible strategies of the players include all positive effort levels not just the ones you show.

The NE (0.25;0.25) is not actually an equilibrium because either player can reduce their effort a tiny amount (but less than the amount in your game) increasing their payoff. For example the optimum amount of effort for player 1 if player 2 uses effort level 0.25 is 0.1875 which is not in your game. If player 1 opts for an effort level of 0.1875 player 2's optimal effort is 3/4*0.1875=0.140625. Continuing this reasoning you'll find eventually there is only one NE: (0;0) which is what you had correctly identified using your second approach.